Tensor products and vector valued functions

Given a non-empty set $S$ and a Banach space $X$. Let $B(S,X)$ be the space of all bounded maps from $S$ to $X$. Can we identify $B(S,X)$ with $\ell^\infty(S) \otimes X$, where $\otimes$ is some kind of tensor product of Banach spaces?

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A tensor product of Banach spaces generally means a completion of the algebraic tensor product under some norm. In this case the algebraic tensor product $\ell^\infty(S) \otimes_a X$ embeds naturally into $B(S,X)$, whose norm is given. So the question becomes, whether $\ell^\infty(S) \otimes_a X$ is dense in $B(S,X)$, and the answer in general is no.
Let $S = \mathbb{N}$ and let $X$ be an infinite dimensional separable Banach space. Choose a sequence $x_n$ which is dense in the unit sphere of $X$. Define $f : S \to X$ by $f(n) = x_n$, so $f \in B(S,X)$. But for any $g \in \ell^\infty(S) \otimes_a X$, the range of $g$ is contained in a finite dimensional subspace $E_g$ of $X$. By Hahn-Banach we can find $x$ in the unit sphere with $d(x, E_g) > 1/2$, and by density there is $x_n$ with $||x_n - x|| < 1/4$. Thus $\sup_n ||g(n) - f(n)|| > 1/4$ so $f$ is not in the closure of $\ell^\infty(S) \otimes X$.
Thus, there is no better description of $B(S,X)$ than $C(\beta S, X)$, where $S$ carries the discrete topology... –  balzac Jun 21 '11 at 18:12
@balzac: This might be a very naive question, but why does $C(\beta S, X) = B(S,X)$? The universal property of $\beta S$ guarantees that continuous functions on $S$ valued in a compact space extend to $\beta S$, but bounded subsets of $X$ need not be precompact. –  Nate Eldredge Jun 21 '11 at 18:53
You're right: I am wrong. Btw, correct me if I am wrong: if $S$ is uncountable then there exist $f\in B(S,\mathbb{R})$ which is not a uniform limit of functions taking finitely many values? –  balzac Jun 21 '11 at 19:19
@balzac: No. Any $f \in B(S, \mathbb{R})$ is a uniform limit of functions taking finitely many values. For instance, say $0 \le f \le 1$; then if $f(x) \in [k/n, (k+1)/n)$, set $f_n(x) = k/n$, and we have $||f - f_n||_\infty \le 1/n$. More generally, this works for any function taking values in a totally bounded metric space. –  Nate Eldredge Jun 21 '11 at 19:40